# For an non-constant and non-vanishing entire function $f$ , the set $U = { z in C : | f(z) | <1 }$ has unbounded connected component.

Suppose $$f$$ is an non constant entire function with $$f(z) neq 0$$ for all $$z in C$$. Then the set $$U = { z in C : | f(z) | <1 }$$ has unbounded connected component.

My attempt

I have concluded some facts:

• $$U$$ is an open set;

• $$f(z) neq 0$$ for all $$z in C$$, then there exists $$epsilon >0$$ such that $$| f(z) | > epsilon$$;

• Since $$f$$ is non-vanishing, we can define another entire function g such that $$g(z) = frac{1}{ f(z) }$$.

But I was not able to conclude the result.

Mathematics Asked by Learner_Shas on January 1, 2021

note that assuming $$f$$ non-constant ($$f=2$$ has $$U$$ empty), $$g=1/f$$ is non-constant and entire hence $$g$$ is unbounded, so $$U$$ being given by $$|g(z)|>1$$ must be unbounded (hence it has at least one unbounded connected component) as $$V$$ given by $$|g(z)| le 1$$ cannot contain the exterior of a disc radius $$R$$ for all $$R$$ large enough as that would imply $$g$$ bounded (as $$g$$ is bounded inside the closed disc of radius $$R$$, if it were bounded by $$1$$ outside, it would be bounded on the plane), while the union of $$U$$ and $$V$$ is the plane! Done!

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